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The x co-ordinate of a point on the curve 9y^2 = x^3 , the normal at which cuts off equal intercept on the axes is ?

Mathematics
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9y^2=x^3 d(9y^2)/dx =d(x^3)/dx 18y dy/dx =3x^2 dy/dx =x^2/(6y)
The gradient of tangent is x^2/(6y) So, gradient of normal is -6y/(x^2)
NOW, let x-intercept of the normal be a then y-intercept is also a.

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Other answers:

Gradient of normal = (a-0)/(0-a) -6y/(x^2) =-1
Now,|dw:1359201825424:dw| and solve for x
|dw:1359201971611:dw|
U need to solve that
|dw:1359202095063:dw|
9y^2=x^3
-6y/(x^2) =-1......then wat did u do ?
Substitute y interms of x

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